SpringBoot统一异常拦截处理(filter中的异常无法被拦截处理)

心已赠人 2022-03-18 07:58 999阅读 0赞

统一异常的拦截处理,网上也是很多资料了,这里贴一下我的异常处理代码

@Slf4j
    @RestControllerAdvice( annotations = RestController.class )
    public class ApiExceptionHandler extends ResponseEntityExceptionHandler {
    
        @ExceptionHandler( BizException.class )
        public ResponseEntity<R<String>> otherException( BizException e ) {
            log.error( "业务异常: {}", e );
    
            R<String> response = R.<String> builder()
                                  .code( RCode.ERROR.getCode() )
                                  .msg( e.getMessage() )
                                  .build();
    
            return new ResponseEntity<>( response, HttpStatus.OK );
        }
    
        @ExceptionHandler( IllegalArgumentException.class )
        public ResponseEntity<R<String>> otherException( IllegalArgumentException e ) {
            log.warn( "非法参数：{}", e );
    
            R<String> response = R.<String> builder()
                                  .code( RCode.INVALID_PARAM.getCode() )
                                  .msg( e.getMessage() )
                                  .build();
    
            return new ResponseEntity<>( response, HttpStatus.OK );
        }
    
        @Override
        protected ResponseEntity<Object> handleHttpMessageNotReadable( HttpMessageNotReadableException ex,
                                                                       HttpHeaders headers, HttpStatus status,
                                                                       WebRequest request ) {
            log.warn( "内部错误：{}", ex.getMessage(), ex );
    
            R<String> response = R.<String> builder()
                                  .code( RCode.ERROR.getCode() )
                                  .msg( ex.getMessage() )
                                  .build();
    
            return new ResponseEntity<>( response, HttpStatus.INTERNAL_SERVER_ERROR );
        }
    }

但是,这个异常处理只能够处理controller中的异常,对于filter中的异常无法捕获,所以我在网上也找了一些,基本上都是继承抽象类或者实现一些接口,比如BasicErrorController/ErrorController,我尝试了一下,未能解决我自己的问题,还是无法拦截filter中的异常,所以我在最后不得已而为之,用了一个不是办法的办法:

那就是直接在filter中将错误信息和错误状态码输出到浏览器,我直接不经过异常处理器,也没有处理异常,而是直接把我想告诉浏览器的信息直接输出.代码如下:

@Override
    	public void doFilter(ServletRequest servletRequest,
    			ServletResponse servletResponse, FilterChain filterChain)
    			throws IOException, ServletException {
    		
    		if (!fasle) {
    			Map<String, String> map = new HashMap<>();
    			map.put("code", "9999");
    			map.put("msg", "没有访问权限，如需要访问，请联系管理员!");
    			response.setCharacterEncoding("utf-8");
    			response.getWriter().print(JSONObject.toJSON(map));
    			return;
    		}
    
    		filterChain.doFilter(servletRequest, servletResponse);
    		return;
    	}

以上完全是抖机灵的做法,还请大佬指点一个能够捕获这个filter中异常的方法!